What exactly does p-wave mean when referring to particles, like charmonia states.

For example, see the following reference:


My doubt is - p wave in scattering means l=1. But how is that used to label a particle state?



1 Answer 1


You're correct that "$p$-wave" in this context means that the charmonium has orbital angular momentum $L=1$. The principal observable effect is that two-particle states with even $L$ have even parity, while those with odd $L$ change sign under parity.

If you were to skim the Particle Data Group's list of charmonia (or the $c\bar c$ section of the short list) you would find that each particle has a value listed for spin, $J$, and parity, $P$. The lightest few states are

  • $\eta_c(1S)$, with $J^P = 0^-$. This is a charm-anticharm bound state where the two quarks have opposite spins and the angular momentum $L=0$. (The total parity is negative because the intrinsic parities of the $c$ and $\bar c$ have opposite sign.)

  • $J/\psi(1S)$, with $J^P=1^-$. This has the same parity as the spinless $\eta_c$, but a unit of spin. So it must be that here the two quarks have the same spin and $L=0$.

  • $\chi_{c0}(1P)$, with $J^P=0^+$. The change in parity suggests that this state must have a different $L$ than the $\eta_c$ and $J/\psi$, and it's the first with a $P$ in its name. There's also a $\chi_{c1}$ and $\chi_{c2}$ with $J^P=1^+, 2^+$, respectively. That suggests these are the three spin combinations of a charm-anticharm pair with spin $S=1$ and orbital angular momentum $L=1$.

There are also some charmonia with $(2S)$ and $(2P)$ in their names. There don't appear to be any definitively labeled $d$-wave charmonia.


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